Question: Let $f(x, y, z) = z + x^2 - y^2$ Suppose $\vec{a} = (-1, 2, 0)$ and $\vec{v} = \left( 1, 2, 1 \right)$. Find the directional derivative of $f(x, y, z)$ at $\vec{a}$ in the direction of $\vec{v}$. Do not normalize the direction vector for your calculation. $\dfrac{\partial f}{\partial v} = $
Explanation: The directional derivative of a function $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $\left( \nabla f (\vec{a}) \right) \cdot \vec{v}$. Let's find the gradient of $f$. $\begin{aligned} \nabla f &= \left( \dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z} \right) \\ \\ &= \left( 2x, -2y, 1 \right) \end{aligned}$ Plugging in $(-1, 2, 0)$, we can find $\nabla f$ at $\vec{a}$. $\nabla f (\vec{a}) = (-2, -4, 1)$ Therefore: $\begin{aligned} \left( \nabla f (\vec{a}) \right) \cdot \vec{v} &= (-2, -4, 1) \cdot (1, 2, 1) \\ \\ &= -2(1) -4(2) + 1(1) \\ \\ &= -9 \end{aligned}$ In conclusion, the directional derivative of $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $-9$.